Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = 3.14159265358979;
const double eps = 1e-7;
const int INF_INT = 0x3f3f3f3f;
const long long INF_LL = 0x3f3f3f3f3f3f3f3f;
const long long mod = (0 ? 1000000007 : 998244353);
const int maxn = 100110;
int n, m, k;
int q;
struct Edge {
  int to, next;
} edge[maxn << 1];
int tot, head[maxn];
void initedge() {
  tot = 0;
  memset(head, -1, sizeof(head));
}
void addedge(int u, int v) {
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot;
  ++tot;
}
struct Path {
  int sz;
  int num[11];
  Path() {
    sz = 0;
    memset(num, 0x3f, sizeof(num));
  }
  void operator+=(int x) {
    num[10] = x;
    sort(num, num + 11);
    sz = min(sz + 1, 10);
  }
  Path operator+(const Path& r) const {
    Path res;
    int i = 0, j = 0, k = 0;
    while (k < 10 && i < sz && j < r.sz) {
      if (num[i] <= r.num[j])
        res.num[k++] = num[i++];
      else
        res.num[k++] = r.num[j++];
    }
    while (i < sz && k < 10) res.num[k++] = num[i++];
    while (j < r.sz && k < 10) res.num[k++] = r.num[j++];
    res.sz = k;
    return res;
  }
  void init() {
    sz = 0;
    memset(num, 0x3f, sizeof(num));
  }
  void print(int k) {
    k = min(sz, k);
    cout << k;
    for (int i = 0; i < k; ++i) {
      cout << ' ' << num[i];
    }
  }
} path[maxn][20];
int fa[maxn][20], dep[maxn];
void dfs(int u) {
  for (int i = 1; (1 << i) <= dep[u]; ++i) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
    path[u][i] = path[u][i - 1] + path[fa[u][i - 1]][i - 1];
  }
  for (int id = head[u], v; id >= 0; id = edge[id].next) {
    v = edge[id].to;
    if (v == fa[u][0]) continue;
    fa[v][0] = u;
    dep[v] = dep[u] + 1;
    dfs(v);
  }
}
Path res;
void getAnce(int& x, int k) {
  for (int i = 0; i < 20; i++) {
    if ((1 << i) & k) {
      res = res + path[x][i];
      x = fa[x][i];
    }
  }
}
void LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  getAnce(u, dep[u] - dep[v]);
  if (u == v) {
    res = res + path[v][0];
    return;
  }
  for (int i = 19; i >= 0; --i) {
    if (fa[u][i] != fa[v][i]) {
      res = res + path[u][i] + path[v][i];
      u = fa[u][i];
      v = fa[v][i];
    }
  }
  res = res + path[u][0] + path[v][1];
  return;
}
int idx[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> q;
  initedge();
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    addedge(u, v);
    addedge(v, u);
  }
  for (int i = 1; i <= m; ++i) {
    cin >> idx[i];
    path[idx[i]][0] += i;
  }
  dep[1] = 1;
  fa[1][0] = 0;
  dfs(1);
  for (int i = 1, u, v, a; i <= q; ++i) {
    res.init();
    cin >> u >> v >> a;
    LCA(u, v);
    int k = min(a, res.sz);
    res.print(k);
    cout << '\n';
  }
  return 0;
}